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Different types of exponential growth equations?

  1. Jul 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A bacteria culture initially contanis 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.
    a. Find an expression for the number of bacteria after t hours.
    b. find the number of bacteria after 3 hours
    c. find the rate of growth after 3 hours.
    d. when will population reach 10,000?​

    2. Relevant equations
    This is done in section of the book of expontial growth and decay, so equations
    dP/dt = kP
    P(t) = P0 e^kt

    3. The attempt at a solutionI
    I managed to solve it, but I had some questions about the exponentials.
    For part a, I got
    P'(t) = 100P, and if P'(1) = 420, P = 4.2

    However, for c., I had to end up taking the differential of P(t) to get P'(t) = 143.5e^1.435t

    Ultimately, at t = 1, these give the same answer. But I don't understand how I'm supposed to know when to get either equation... Would my answer for C have been incorrect for A?
  2. jcsd
  3. Jul 5, 2015 #2
    The answer for C is not the same as for A.

    They can't be, they have different units. Take care with your units in each part. P'(t) = 143.5e^1.435t needs units. What are they?
  4. Jul 5, 2015 #3
    One is a rate (cells/hour), one is just a population (number of cells). It appears I didn't word my question right,. Thank you for your help so far.

    But if A is a population equation, and C is a change in rate of population growth, why wouldn't my answer to A have been P(t) = P0e^(kt) form? My book used 100(4.2)^t for (a) and it didn't make sense to me. Is there a way that population rate of growth could've been found from that? Or were they trying to draw some comparison?
  5. Jul 6, 2015 #4
    There are two ways of modeling exponential growth: P(t) = Ab^t, where A and b are constants and P(t) = Aexp(kt), where A and k are constants.

    The way you did it is more common for bacterial growth and any time one solves the same differential equation you posted. One can use simple rules of logs and exponents to shot that the two models are equivalent and to compute the equivalent model in the other form when given one to start with.
  6. Jul 9, 2015 #5
    Thank you for your help! I understand now.
  7. Jul 9, 2015 #6


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    You can use both [itex]Aexp(kt)= Ae^{kt}[/itex] and [itex]Ab^t[/itex] for the same thing because [itex]e^x[/itex] and [itex]ln(x)[/itex] are "inverse functions": [itex]e^{ln(x)}= x[/itex]. In particular [itex]p^t= e^{ln(p^t)}= e^{t ln(p)}= e^{kt}[/itex] where [itex]k= ln(p)[/itex].
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